Answer:
B: increase.
Explanation:
When we are considering two gases A and B in a container at room temperature .
We have to find the change on rate of reaction when the number of molecules of gases A is doubled
Let [A]=a and [B]=b
A+B
product
Rate of reaction
![R_1=k[A][B]=kab](https://tex.z-dn.net/?f=R_1%3Dk%5BA%5D%5BB%5D%3Dkab)
We know that concentration is increases with increase in number of moles
When the number of molecules of gases A is doubled then concentration of gases A increases.
Therefore ,[A]=2a
Rate of reaction
Hence, the rate of reaction is 2 times the initial rate of reaction.Therefore, the rate of reaction will increase when the number of molecules of gases A is doubled.
Answer: B: increase.
Answer:
3.1 kg
Explanation:
Step 1: Write the balanced combustion equation
C₈H₁₈ + 12.5 O₂ ⇒ 8 CO₂ + 9 H₂O
Step 2: Calculate the moles corresponding to 1.0 kg of C₈H₁₈.
The molar mass of C₈H₁₈ is 114.23 g/mol.
1.0 × 10³ g × 1 mol/114.23 g = 8.8 mol
Step 3: Calculate the moles of CO₂ produced from 8.8 moles of C₈H₁₈
The molar ratio of C₈H₁₈ to CO₂ is 1:8. The moles of CO₂ produced are 8/1 × 8.8 mol = 70 mol.
Step 4: Calculate the mass corresponding to 70 moles of CO₂
The molar mass of CO₂ is 44.01 g/mol.
70 mol × 44.01 g/mol = 3.1 × 10³ g = 3.1 kg
The balanced chemical equation is given as:
2CH3CH2OH(l) → CH3CH2OCH2CH3(l) + H2O(l)
We are given the yield of CH3CH2OCH2CH3 and the amount of ethanol to be used for the reaction. These values will be the starting point for the calculations.
Theoretical amount of product produced:
329 g CH3CH2OH ( 1 mol / 46.07 g ) ( 1 mol CH3CH2OCH2CH3 / 2 mol CH3CH2OH ) (74.12 g / mol ) = 264.66 g CH3CH2OCH2CH3
% yield = .775 = actual yield / 264.66
actual yield = 205.11 g CH3CH2OCH2CH3
<span>STP means standard temperature
and pressure at 0°C (273K) and 1 atm (atmosphere). The density of the unknown
gas is 0.63 gram per liter. The deal gas equation is PV = nRT. The n is the
numer of moles and can be represented as mass of the gas, m, divided by the
molar mass, c. so we have,</span>
PV = nRT
PV = (m/c)RT
Since the density is d = m/V
Pc = (m/V)RT
Pc = dRT
c = drT/P
substitute the values into the equation,
c = [(0.63g/L)(0.08206
L-atm/mol-K)(273K)]/(1atm)
<u>c = 14.11 g/mol</u>
<span><span>When you write down the electronic configuration of bromine and sodium, you get this
Na:
Br: </span></span>
<span><span />So here we the know the valence electrons for each;</span>
<span><span>Na: (2e)
Br: (7e, you don't count for the d orbitals)
Then, once you know this, you can deduce how many bonds each can do and you discover that bromine can do one bond since he has one electron missing in his p orbital, but that weirdly, since the s orbital of sodium is full and thus, should not make any bond.
However, it is possible for sodium to come in an excited state in wich he will have sent one of its electrons on an higher shell to have this valence configuration:</span></span>
<span><span /></span><span><span>
</span>where here now it has two lonely valence electrons, one on the s and the other on the p, so that it can do a total of two bonds.</span><span>That's why bromine and sodium can form </span>
<span>
</span>
